Step 1: Picture the situation.
An electron whirls around a tiny circular orbit. A moving charge that keeps going round and round behaves like a small current loop, and every current loop has a magnetic moment. We must find that moment.
Step 2: Collect the data in SI units.
Radius $r = 0.05\ \text{nm} = 5 \times 10^{-11}\ \text{m}$, revolution frequency $f = 10^{14}\ \text{s}^{-1}$, and charge $e = 1.6 \times 10^{-19}\ \text{C}$.
Step 3: Treat the orbiting electron as a current.
If a charge $e$ passes a point $f$ times per second, the equivalent current is $I = e f = (1.6 \times 10^{-19})(10^{14}) = 1.6 \times 10^{-5}\ \text{A}$.
Step 4: Find the area enclosed by the orbit.
The loop area is $A = \pi r^2 = \pi (5 \times 10^{-11})^2 = 25\pi \times 10^{-22}\ \text{m}^2$.
Step 5: Combine into the magnetic moment.
The magnetic moment of a current loop is $M = I A$. So $M = (1.6 \times 10^{-5})(25\pi \times 10^{-22}) = 40\pi \times 10^{-27}\ \text{A}\cdot\text{m}^2$.
Step 6: Evaluate the number.
$40\pi \times 10^{-27} = 4\pi \times 10^{-26} \approx 1.26 \times 10^{-25}\ \text{A}\cdot\text{m}^2$, which matches the option carrying the $1.26$ coefficient. So the answer is option (4).
\[ \boxed{M = 4\pi \times 10^{-26}\ \text{A}\cdot\text{m}^2 \approx 1.26 \times 10^{-25}\ \text{A}\cdot\text{m}^2} \]